I`m stuck in what seems like a very basic derivation.
I`m having my very first contact with the covariant formulation of Electrodynamics, and I want to show that $$\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma} = 4 \vec{E}\cdot{B}$$
but I don`t know how to begin. I've found two posts about the subject of Lorentz invariants of the electromagnetic tensor(namely this one and this one) but it didn't acttually help me
I appreciate any guiding thoughts that help me get it started.
With $\epsilon^{\mu \nu \rho \sigma}$ equal to 1 for even permutations, -1 for odd permutations of the indices, 0 with repeated indices, and remembering that
$$F_{\mu \nu} = \begin{bmatrix} 0 & e_x/c & e_y/c & e_z/c \\ -e_x/c & 0 & -b_z & b_y \\ -e_y/c & b_z & 0 & -b_x \\ -e_z/c & -b_y & -b_y & 0 \\ \end{bmatrix} \ ,$$
using Cartesian coordinates for space, direct computation of the desired invariant is the sum of 24 terms (here only 6 of them are shown, those with permutations that start with 0; all the sums of permutations starting with other indices give the same result - check it!) that reads
$$\begin{aligned} \epsilon^{\mu \nu \rho \sigma} F_{\mu \nu} F_{\rho \sigma} & = \epsilon^{0123} F_{01} F_{23} + \epsilon^{0132} F_{01} F_{32} + \epsilon^{0231} F_{02} F_{31} + \epsilon^{0213} F_{02} F_{13} + \epsilon^{0321} F_{03} F_{21} + \epsilon^{0312} F_{03} F_{12} + ... = \\ & = \frac{e_x}{c} \left( 1 \cdot (- b_x) - 1 \cdot b_x \right) + \frac{e_y}{c} \left( 1 \cdot (- b_y) - 1 \cdot b_y \right) + \frac{e_z}{c} \left( 1 \cdot (- b_z) - 1 \cdot b_z \right) + ... = \\ & = - \frac{2}{c} \left( e_x b_x + e_y b_y + e_z b_z \right) + \text{other 3 times} =\\ & = - \frac{8}{c} \vec{e} \cdot \vec{b} \ , \end{aligned}$$
so you're off by a factor 2 and a sign, w.r.t. the results you should get using $(+,-,-,-)$ and common definition of the electromagnetic tensor.
There are 24 non-zero components for the fully anti-symmetric pseudo-tensor $ε^{μνρσ}$, with $ε^{0123} = 1$. Since $$ε^{μνρσ}F_{μν}F_{ρσ} = ε^{νμρσ}F_{νμ}F_{ρσ} = ε^{μνσρ}F_{μν}F_{σρ} = ε^{ρσμν}F_{ρσ}F_{μν},$$ owing to the anti-symmetry of $F_{μν}$, and to $ε^{μνρσ} = ε^{ρσμν}$ (which is a consequence of the anti-symmetry of $ε^{μνρσ}$) then the sum arranges itself into 3 octuplets as: $$ε^{μνρσ}F_{μν}F_{ρσ} = 8\left(F_{01}F_{23} + F_{02}F_{31} + F_{03}F_{12}\right).$$
By convention, in SI, $A_0 = -φ$, the electric (or scalar) potential, $\left(A_1, A_2, A_3\right) = $, the magnetic (or vector) potential, the electric and magnetic fields are given in terms of the potential by $$ = ∇×,\quad = -\frac{∂}{∂t} - ∇φ,$$ and $F_{μν} = ∂_μA_ν - ∂_νA_μ$, $∂_0 = ∂/∂t$, $\left(∂_1, ∂_2, ∂_3\right) = ∇$, so $$\left(F_{01}, F_{02}, F_{03}\right) = \frac{∂}{∂t} + ∇φ = -,\quad \left(F_{23}, F_{31}, F_{12}\right) = ∇× = .$$ Therefore $$ε^{μνρσ}F_{μν}F_{ρσ} = 8\left(F_{01}F_{23} + F_{02}F_{31} + F_{03}F_{12}\right) = -8·.$$
There is no universal consensus on the conventions used for indexing, nor even for the scaling of the fields and their components. Therefore, you can see a diversity of answers, depending on what convention is used. But, this is the representation that most closely matches what's in Maxwell's treatise and (therefore) the SI.
To give you an example of how the answer may vary: if instead of indexing $x^0 = t$, you index the coordinate as $x^0 = ct$, where $c$ denotes light speed, then this leads to the changes, $A_0 = -φ/c$, $∂_0 = (1/c)∂/∂t$, $\left(F_{01},F_{02},F_{03}\right) = /c$ and (thus) to: $ε^{μνρσ}F_{μν}F_{ρσ} = -8·/c$.
